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Hard Time Understanding Archimedes' Principle.

  1. Oct 19, 2012 #1
    If I have a body (for simplicity a cube with side d) on a fluid (with density ρ) with its top at a depth h, the force acting on top of it is ma = (h*d*d)*ρ*(g); where g is the acceleration of gravity, (d*d*h) is the volume of the column of liquid on top of the cube and that times ρ is the mass of the column. (Assume no atmospheric pressure).

    Now, I get everything up to that point. What I don't get is the force acting on the bottom of the cube. Shouldn't the force being exerted by the column of liquid bellow the cube on the cube be equal and opposite to the weight of the column of liquid on top of the cube plus the weight of the cube? But the derivation I saw said that the force acting on the bottom is equal to ((h + d)*d*d)*ρ*(g); where ((h + d)*d*d) is the volume of the column of liquid on top of the cube plus the volume of the cube. Isn't that assuming that the density of the cube is the same as the density of the liquid?

    I don't get how you can get to Archimedes principle without assuming that the net force on the cube (buoyant force), is equal to the volume of liquid displaced by the cube in the first place.
  2. jcsd
  3. Oct 20, 2012 #2


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    Hi V0ODO0CH1LD! :smile:

    The pressure at depth x has nothing to do with what is in the way, so long as there's a way round it.

    It is always ρgx (or ∫ρgdx if ρ is not constant), and then you multiply by the surface area to get the force.

    (and remember of course that the body is not in equilibrium)

    If the object completely blocks the way … eg a tube with water, then a body with the same diameter as the tube, then more water … then of course you must use the actual mass of the body, not the mass of the displaced water. :wink:
  4. Oct 20, 2012 #3
    Consider a column of liquid just to the right of left of the cube. The pressure in this column at the depth of the base of the cube is ρg(h+d). This must also be the pressure directly underneath the base of the cube, or else liquid would be flowing horizontally from the higher pressure region to the lower pressure region. But there is no flow. So the liquid force on the bottom of the cube must be ρg(h+d)d2. The net upward force of the liquid pressure on the cube must be ρgd3, which is just the weight of the liquid that would have been present if the cube were filled with liquid. Another way of thinking about this is that the liquid doesn't know whether the space occupied by the cube contains a material of a different density or whether it still contains liquid. As far as it knows, there is still liquid there. So it exerts an upward force as if there were still liquid there.
  5. Oct 20, 2012 #4
    I still don't get the difference between the tube tiny tim described and just the object sitting on an ocean filled with liquid (in my mind the only things interacting with the cube are the columns of water bellow and above it; is that wrong?).
    I get that it is the way you guys are describing, I just don't understand why.
    Last edited: Oct 20, 2012
  6. Oct 20, 2012 #5


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    no, you're basically correct, the only vertical interaction with the cube is the water in contact with its upper and lower surfaces

    but the important point is that the vertical interactions with the water below the cube include force transmitted through water round the cube

    (and in the tube example, there is no such water!)

    you can squeeze a plum without putting much pressure on the stone …

    the plum sort of flows round the stone, and changes shape

    the stone isn't in the way​

    ok, water doesn't actually change shape (if it's in a jar, say), but force does get transmitted through it (missing the stone) in the same way :wink:
  7. Oct 20, 2012 #6


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    No because if the cube was weightless the force on the bottom would be same as the top. That can't be true because it's deeper so the pressure is higher.

    You are right but it's better to think of it as...

    (fluid pressure at a depth = h+d) * the area * ...
  8. Oct 20, 2012 #7
    No. That would only be true if the cube were in equilibrium, which it is not. It would stay put only if you held it in place. Otherwise it would rise or drop. Your holding it in place would put an additional force on the combination of cube and column, so that the force on the bottom would not be the sum of the weights of the column and cube.
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